A particle moves in xy plane with a velocity given by `vecV = (8t-2)hati+2hatj.`. If it passes through the point `(14,4)` at `t=2 sec , then give eqution of the point

A particle moves in the x-y plane with velocity v_x = 8t-2 and v_y = 2. If it passes through the point x =14 and y = 4 at t = 2 s, the equation of the path is

A particle moves in plane with acceleration veca = 2hati+2hatj . If vecv=3hati+6hatj , then final velocity of particle at time t=2s is

A particle A moves with velocity (2hati-3hatj)m//s from a point (4,5m)m . At the same instant a particle B , moving in the same plane with velocity (4hati+hatj)m//s passes through a point C(0,-3)m . Find the x -coordinate (in m ) of the point where the particles collide.

A particle moves in the xy plane and at time t is at the point (t^(2), t^(3)-2t) . Then :-

A particle moves in xy plane with its position vector changing with time (t) as vec(r) = (sin t) hati + (cos t) hatj ( in meter) Find the tangential acceleration of the particle as a function of time. Describe the path of the particle.

Find the vector and cartesian equations of the planes : (a) that passes through the point (1,0,-2) and the normal to the plane is hati + hatj - hatk (b) that passes through the point (1,4,6) and the normal vector of the plane is hati - 2 hatj + hatk .

An ant is moving in x-y plane with its position vector vecs=2t^2hati+5hatj then, velocity vector of it at t = 1 sec is:

A particle moves in xy-plane. The position vector of particle at any time is vecr={(2t)hati+(2t^(2))hatj}m. The rate of change of theta at time t = 2 second. ( where 0 is the angle which its velocity vector makes with positive x-axis) is :